A current element 50dl  is at (0,0,0)along x-axis.If dl is 10 raised to -2 m find the magnetic field at a distance of 0.25m on y-axis.Calculate the strength of the field.

Dear Student,

Current element i am considering as 5 dl as mentioned in your comment next to the question.

Please find below the solution to the asked query:

According to Biot - Savart Law, the equation for the magnetic field due to a current element at a distance d is,

$B=\frac{{\mu }_0}{4\pi }\frac{Idl}{d^2}$
Here, the distance is d = 0.25 m, length of the current element is $Idl=5\times {10}^{-2}=0.05 Am$.

Therefore, the magnetic field is,

$$\begin{gathered} B=\frac{{\mu }_0}{4\pi }\frac{Idl}{d^2}=\frac{4\pi \times {10}^{-7}}{4\pi }\times \frac{0.05}{0.25} \\ \Rightarrow B=0.2\times {10}^{-7} \\ \Rightarrow B=20 nT \end{gathered}$$

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